Notation - questions like - How much more probable is getting 5 heads and 5 tails than 9 heads and 1 tail?

Question

Original Question: A well-balanced coin is flipped 10 times. How much more likely is it to get 5Heads & 5Tails vs 9Heads & 1Tail?

The math operations aren't confusing for me. I just have problems thinking through the problems, which is more a notation-related problem, understanding what the math labels.

A fair coin has an equal chance of hitting heads || tails. ie. 50% likelihood of anything.

In general for problems like this:

the formula = multiplicity of event x probability associated with sub-events.

In this case, multiplicity is determined by the combination formula, because the desired outcome is content specific, not priority specific.

Ω = multiplicity

Ω_5H = C(10,5)=252

Ω_9H = C(10,9)=C(10,1)=10

(1/2)^10 <-- associated with flipping coin 10 times.

What specifically does the (1/2)^10 describe in a generic sense?

• The probability of all subevents?
• What do you call the situation of flipping the coin 10 times?

in assessing the likelihood ratio: it's the same if you just consider the multiplicities.

Answer 1

Problems like these are group events. This is why you need to consider multiplicities and why a probability factor is scaled.

Finding the likelihood of getting 5Heads & 5Tails || 9Heads & 1Tail ... is actually a problem of finding the probability for a set of independent events.

In this description getting heads on one flip assumes that getting heads on the next flip is also equally random.

Thus going back to the formula:

the probability of the group event || set of independent events is defined as thus

Pgroup= Ωgroup * (probability of individual event)^number of component events.

Ωgroup= the number of multiple ways you can acquire the group arrangement

probability of individual event = defined by some process described in the problem

Going back to the example problem:

• P_specific_combination= Ωcombination x (probability of getting head on 1 flip)^#flips

• P_5H,5T = C(10,5) x (1/2)^10

• (1/2)^10 describes the getting heads on each flip.

Answer 2

The "sub-events" you are talking about are usually called outcomes. These are individual elements of the sample space (as opposed to events, which are subsets of the sample space).

So the sample space of flipping a coin $10$ times is the set of all ordered strings of $H,T$ of length $10$: $$S=\{(x_1,x_2,\dots,x_{10}) : x_i = H \text{ or T}, 1 \leq i \leq 10\}.$$ An outcome of this sample space is an individual string, e.g., $(H,T,H,H,T,T,T,H,H,T)$. Each outcome in this sample space is associated with a probability, which is $(1/2)^{10}$ (i.e., all outcomes are equally likely).

An event in this sample space could be something like the event of getting $9$ heads and $1$ tail. This event is a set consisting of $10$ different outcomes: $$\{(H,H,H,H,H,H,H,H,H,T), (H,H,H,H,H,H,H,H,T,H), \dots\}.$$ The probability of an event is the sum of the probabilities of all outcomes in the event. Since all outcomes are equally likely with probability $(1/2)^{10}$, this works out to $(\#\text{outcomes in the event} ) \times(1/2)^{10}$.

The situation of flipping the coin $10$ times could be called different things depending on the context. For instance, the general setup could be called an "experiment" ("consider the experiment of flipping a fair coin ten times..."). An particular instance of this experiment would be called an outcome as I explained above.